Linear Equations Worksheets with Answer Key - Free Printable

Solving Linear Equations Worksheet

We will solve each equation step by step. Let's go through them one by one.

---

#### Problem 1: \( 2(m + 6) = 48 \)

1. Distribute the 2 on the left-hand side:
\[
2(m + 6) = 2m + 12
\]
So the equation becomes:
\[
2m + 12 = 48
\]

2. Subtract 12 from both sides to isolate the term with \( m \):
\[
2m + 12 - 12 = 48 - 12
\]
\[
2m = 36
\]

3. Divide both sides by 2 to solve for \( m \):
\[
\frac{2m}{2} = \frac{36}{2}
\]
\[
m = 18
\]

Solution:
\[
\boxed{18}
\]

---

#### Problem 2: \( 8x + 9 - 3x = 8 + 4x + 5 \)

1. Combine like terms on both sides:
- Left-hand side: \( 8x - 3x + 9 = 5x + 9 \)
- Right-hand side: \( 8 + 5 + 4x = 13 + 4x \)
So the equation becomes:
\[
5x + 9 = 13 + 4x
\]

2. Subtract \( 4x \) from both sides to get all \( x \)-terms on one side:
\[
5x - 4x + 9 = 13 + 4x - 4x
\]
\[
x + 9 = 13
\]

3. Subtract 9 from both sides to isolate \( x \):
\[
x + 9 - 9 = 13 - 9
\]
\[
x = 4
\]

Solution:
\[
\boxed{4}
\]

---

#### Problem 3: \( 8a - (4a + 32) = 16 \)

1. Distribute the negative sign on the left-hand side:
\[
8a - 4a - 32 = 16
\]

2. Combine like terms:
\[
4a - 32 = 16
\]

3. Add 32 to both sides to isolate the term with \( a \):
\[
4a - 32 + 32 = 16 + 32
\]
\[
4a = 48
\]

4. Divide both sides by 4 to solve for \( a \):
\[
\frac{4a}{4} = \frac{48}{4}
\]
\[
a = 12
\]

Solution:
\[
\boxed{12}
\]

---

#### Problem 4: \( -3(x - 1) + 8(x - 3) = 6x + 7 - 5x \)

1. Distribute the terms on the left-hand side:
- For \( -3(x - 1) \): \( -3x + 3 \)
- For \( 8(x - 3) \): \( 8x - 24 \)
So the left-hand side becomes:
\[
-3x + 3 + 8x - 24
\]
Combine like terms:
\[
(-3x + 8x) + (3 - 24) = 5x - 21
\]

2. Simplify the right-hand side:
\[
6x + 7 - 5x = (6x - 5x) + 7 = x + 7
\]

3. The equation now is:
\[
5x - 21 = x + 7
\]

4. Subtract \( x \) from both sides to get all \( x \)-terms on one side:
\[
5x - x - 21 = x + 7 - x
\]
\[
4x - 21 = 7
\]

5. Add 21 to both sides to isolate the term with \( x \):
\[
4x - 21 + 21 = 7 + 21
\]
\[
4x = 28
\]

6. Divide both sides by 4 to solve for \( x \):
\[
\frac{4x}{4} = \frac{28}{4}
\]
\[
x = 7
\]

Solution:
\[
\boxed{7}
\]

---

#### Problem 5: \( 10x + 5(5x - 3) = 5(6x + 4) \)

1. Distribute the terms on both sides:
- For \( 5(5x - 3) \): \( 25x - 15 \)
- For \( 5(6x + 4) \): \( 30x + 20 \)
So the equation becomes:
\[
10x + 25x - 15 = 30x + 20
\]

2. Combine like terms on the left-hand side:
\[
(10x + 25x) - 15 = 35x - 15
\]
So the equation is:
\[
35x - 15 = 30x + 20
\]

3. Subtract \( 30x \) from both sides to get all \( x \)-terms on one side:
\[
35x - 30x - 15 = 30x - 30x + 20
\]
\[
5x - 15 = 20
\]

4. Add 15 to both sides to isolate the term with \( x \):
\[
5x - 15 + 15 = 20 + 15
\]
\[
5x = 35
\]

5. Divide both sides by 5 to solve for \( x \):
\[
\frac{5x}{5} = \frac{35}{5}
\]
\[
x = 7
\]

Solution:
\[
\boxed{7}
\]

---

#### Problem 6: \( 0.02m + 0.08(8 - m) = 1.78 \)

1. Distribute the 0.08 on the left-hand side:
\[
0.02m + 0.08 \cdot 8 - 0.08m = 1.78
\]
\[
0.02m + 0.64 - 0.08m = 1.78
\]

2. Combine like terms:
\[
(0.02m - 0.08m) + 0.64 = -0.06m + 0.64
\]
So the equation becomes:
\[
-0.06m + 0.64 = 1.78
\]

3. Subtract 0.64 from both sides to isolate the term with \( m \):
\[
-0.06m + 0.64 - 0.64 = 1.78 - 0.64
\]
\[
-0.06m = 1.14
\]

4. Divide both sides by -0.06 to solve for \( m \):
\[
\frac{-0.06m}{-0.06} = \frac{1.14}{-0.06}
\]
\[
m = -19
\]

Solution:
\[
\boxed{-19}
\]

---

#### Problem 7: \( \frac{1}{3}(10 - 2x) = \frac{1}{3}(8 - 2x) \)

1. Since the denominators are the same, we can equate the numerators directly:
\[
10 - 2x = 8 - 2x
\]

2. Add \( 2x \) to both sides:
\[
10 - 2x + 2x = 8 - 2x + 2x
\]
\[
10 = 8
\]

This is a contradiction, which means there is no solution.

Solution:
\[
\boxed{\text{No solution}}
\]

---

#### Problem 8: \( -14 + 6b + 7 - 2b = 1 + 5b \)

1. Combine like terms on the left-hand side:
- Combine \( b \)-terms: \( 6b - 2b = 4b \)
- Combine constants: \( -14 + 7 = -7 \)
So the left-hand side becomes:
\[
4b - 7
\]
The equation is:
\[
4b - 7 = 1 + 5b
\]

2. Subtract \( 4b \) from both sides to get all \( b \)-terms on one side:
\[
4b - 4b - 7 = 1 + 5b - 4b
\]
\[
-7 = 1 + b
\]

3. Subtract 1 from both sides to isolate \( b \):
\[
-7 - 1 = 1 + b - 1
\]
\[
-8 = b
\]

Solution:
\[
\boxed{-8}
\]

---

Final Answers:

1. \( \boxed{18} \)
2. \( \boxed{4} \)
3. \( \boxed{12} \)
4. \( \boxed{7} \)
5. \( \boxed{7} \)
6. \( \boxed{-19} \)
7. \( \boxed{\text{No solution}} \)
8. \( \boxed{-8} \)